i'm truly sorry if this is the wrong forum, as i'm teaching myself this stuff, i'm unsure of what it's sub forum should be. since my last question was moved here, i put this one here too.

anyway.....

use the following procedure to find the leats ( perpendicular ) distance of the pointfrom the line( 1,2 ), without having to find the equation of a line perpendicular toy = 3x + 5.y = 3x + 5

(a) letbe a general point on the line, Show that its distance,( x,y ), fromdis given by( 1,2 ).d^2 = ( x - 1 )^2 + ( y - 2 )^2

(b) use the equation of the line to show that. I CAN SEE THE ANSWER HERE SO NO WORRIES.d^2 = ( x - 1 )^2 + ( 3x 3 )^2

(c) by completing the square, show that the minimum distance required 3/5√10